The perimiter of a lot is 190 meters. The width is one-fourth of the length. Find the dimensions.
Here's my system of equations:
2w + 2L = 190
1/4L = w
What would I do from here? I prefer the elimination method. Would I multiply the second equation by 4? When I did so, it kind of messed up the equation.
Thanks! :)
I would use substitution, but if you insist on using elimination, multiplying the second equation by 4 gives you
L = 4W or 4W - L = 0
now double that:
8W - 2L = 0 , now the first:
2W + 2L = 190, now add them
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I'm sure you can finish it.
Did you get w = 19 and L = 76?
Thanks! :)
yes, correct!
You are correct in setting up the system of equations, which is a good first step. To solve this system of equations using the elimination method, you need to eliminate one variable.
In this case, to eliminate the fraction in the second equation, you can multiply both sides of the equation by 4. This will give you:
4 * (1/4L) = 4w
L = 4w
Now, you have two equations:
2w + 2L = 190
L = 4w
Since L is already isolated in the second equation, let's substitute this value into the first equation:
2w + 2(4w) = 190
2w + 8w = 190
10w = 190
w = 19
Now that you have the value of w, you can substitute it back into the second equation to find L:
L = 4w
L = 4(19)
L = 76
So, the dimensions of the lot are width (w) = 19 meters and length (L) = 76 meters.